Problem 5

Exercise 1.5 (Weak convergence is finite)   Suppose $\{x_n\}\subseteq X$ converges weakly to $x_0\in X$. Then $\Vert x_0\Vert\leq \liminf \Vert x_n\Vert$.

Proof. Recall that $\lim \phi(x_n)$ exists for each $\phi$, so we may replace the limit with its $\liminf$ and proceed in the double dual

$$\Vert x_0\Vert = \Vert x_0^*\Vert = \sup_{\Vert\phi\Vert=1} \Vert x_0^*(\phi)\Vert = \sup_\phi \vert\phi(x_0)\vert = \sup_\phi \lim_n \vert\phi(x_n)\vert$$ (1.46)

$$= \sup_{\phi} \sup_{n\geq 1} \inf_{m\geq n} \vert\phi(x_m)\vert = \sup_{n\geq 1} \sup_{\phi} \inf_{m\geq n} \vert\phi(x_m)\vert$$ (1.47)

$$\leq \sup_{n\geq 1} \inf_{m\geq n} \sup_{\Vert\phi\Vert=1} \vert\... ...to\infty} \Vert\phi\Vert\Vert x_n\Vert \leq \liminf_{n\to\infty} \Vert x_n\Vert$$ (1.48)

$\qedsymbol$